Apply Spark Column operations directly on a Series

On issue #1492, I noticed it was noted the following operation was possible on a column of a Koalas DataFrame:

>>> import databricks.koalas as ks
>>> df = ks.DataFrame(["example"], columns=["column"])
>>> from pyspark.sql import functions as F
>>> df["column"] = F.trim(F.upper(F.col("column")))
>>> df
    column
0  EXAMPLE

By any chance, is there currently a way to apply Spark SQL/Column functions to a Koalas Series? I can imagine it looking something like so:

>>> import databricks.koalas as ks
>>> import pyspark.sql.functions as F
>>> kss = ks.Series(["example"])
>>> kss.apply(F.trim(F.upper(kss.spark_column)))
0    EXAMPLE
Name: 0, dtype: object

[HyukjinKwon]

tada

>>> kss.spark.transform(lambda s: tokenize(s))
0            [1, example]
1    [2, string, example]
Name: 0, dtype: object

[HyukjinKwon]

I opened a PR. The idea is basically to collect everything related to Spark itself to .spark namespace. This specific issue could be resolved via:

>>> import databricks.koalas as ks
>>> import pyspark.sql.functions as F
>>> kss = ks.Series(["example"])
>>> kss.spark.transform(lambda spark_column: F.trim(F.upper(spark_column)))
0    EXAMPLE
Name: 0, dtype: object

Given that the current APIs in Pandas such as Series.transform, Series.apply, DataFrame.apply, DataFrame.transform, etc, I named it Series.spark.transform and made the usage similar. Also, it shares the same limitation.